Integrand size = 45, antiderivative size = 178 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a (A+7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (24 A+28 B+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a (24 A+28 B+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \] Output:
2/35*a*(A+7*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/105* a*(24*A+28*B+35*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4/ 105*a*(24*A+28*B+35*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2 )+2/7*A*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d*x+c)^(5/2)
Time = 0.87 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \left (15 A+3 (6 A+7 B) \sec (c+d x)+(24 A+28 B+35 C) \sec ^2(c+d x)+(48 A+56 B+70 C) \sec ^3(c+d x)\right ) \sin (c+d x)}{105 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2 ))/Sec[c + d*x]^(7/2),x]
Output:
(2*a*(15*A + 3*(6*A + 7*B)*Sec[c + d*x] + (24*A + 28*B + 35*C)*Sec[c + d*x ]^2 + (48*A + 56*B + 70*C)*Sec[c + d*x]^3)*Sin[c + d*x])/(105*d*Sec[c + d* x]^(5/2)*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.96 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4574, 27, 3042, 4503, 3042, 4292, 3042, 4291}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {\sec (c+d x) a+a} (a (A+7 B)+a (4 A+7 C) \sec (c+d x))}{2 \sec ^{\frac {5}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x) a+a} (a (A+7 B)+a (4 A+7 C) \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)}dx}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (A+7 B)+a (4 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{5} a (24 A+28 B+35 C) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 (A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (24 A+28 B+35 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 (A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {\frac {1}{5} a (24 A+28 B+35 C) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (24 A+28 B+35 C) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {\frac {2 a^2 (A+7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {1}{5} a (24 A+28 B+35 C) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec [c + d*x]^(7/2),x]
Output:
(2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((2 *a^2*(A + 7*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d* x]]) + (a*(24*A + 28*B + 35*C)*((2*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]] *Sqrt[a + a*Sec[c + d*x]]) + (4*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sq rt[a + a*Sec[c + d*x]])))/5)/(7*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 2.82 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.61
| method | result | size |
| default | \(\frac {2 \left (\left (15 \cos \left (d x +c \right )^{3}+18 \cos \left (d x +c \right )^{2}+24 \cos \left (d x +c \right )+48\right ) A +\left (21 \cos \left (d x +c \right )^{2}+28 \cos \left (d x +c \right )+56\right ) B +\left (35 \cos \left (d x +c \right )+70\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(109\) |
| parts | \(\frac {2 A \left (5 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )+16\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{35 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (6 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )+16 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \left (2 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(196\) |
Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2 ),x,method=_RETURNVERBOSE)
Output:
2/105/d*((15*cos(d*x+c)^3+18*cos(d*x+c)^2+24*cos(d*x+c)+48)*A+(21*cos(d*x+ c)^2+28*cos(d*x+c)+56)*B+(35*cos(d*x+c)+70)*C)*(a*(1+sec(d*x+c)))^(1/2)/(c os(d*x+c)+1)/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(7/2),x, algorithm="fricas")
Output:
2/105*(15*A*cos(d*x + c)^4 + 3*(6*A + 7*B)*cos(d*x + c)^3 + (24*A + 28*B + 35*C)*cos(d*x + c)^2 + 2*(24*A + 28*B + 35*C)*cos(d*x + c))*sqrt((a*cos(d *x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))
Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x +c)**(7/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 609 vs. \(2 (154) = 308\).
Time = 0.30 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.42 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(7/2),x, algorithm="maxima")
Output:
1/840*(3*sqrt(2)*(105*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 35*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), c os(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 7*cos(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 105*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 35* cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7 /2*c))) - 7*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos (7/2*d*x + 7/2*c))) + 10*sin(7/2*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2* d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35*sin(3/7*arctan2(sin(7/2*d*x + 7/ 2*c), cos(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), c os(7/2*d*x + 7/2*c))))*A*sqrt(a) + 14*sqrt(2)*(30*cos(4/5*arctan2(sin(5/2* d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*cos(2/5*arct an2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 30 *cos(5/2*d*x + 5/2*c)*sin(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), co s(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2* d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/ 2*c), cos(5/2*d*x + 5/2*c))))*B*sqrt(a) + 140*sqrt(2)*(3*cos(2/3*arctan2(s in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*cos(3 /2*d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2...
Time = 1.31 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 175 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 245 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (147 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 119 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 175 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (27 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 49 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}} d} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c )^(7/2),x, algorithm="giac")
Output:
2/105*(105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*C*a^4*sgn(cos(d*x + c)) + (105*sqrt(2)*A*a^4*sgn(cos( d*x + c)) + 175*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 245*sqrt(2)*C*a^4*sgn(co s(d*x + c)) + (147*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 119*sqrt(2)*B*a^4*sgn (cos(d*x + c)) + 175*sqrt(2)*C*a^4*sgn(cos(d*x + c)) + (27*sqrt(2)*A*a^4*s gn(cos(d*x + c)) + 49*sqrt(2)*B*a^4*sgn(cos(d*x + c)) + 35*sqrt(2)*C*a^4*s gn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2* d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2) *d)
Time = 17.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (420\,A\,\sin \left (c+d\,x\right )+490\,B\,\sin \left (c+d\,x\right )+560\,C\,\sin \left (c+d\,x\right )+126\,A\,\sin \left (2\,c+2\,d\,x\right )+36\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+112\,B\,\sin \left (2\,c+2\,d\,x\right )+42\,B\,\sin \left (3\,c+3\,d\,x\right )+140\,C\,\sin \left (2\,c+2\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \] Input:
int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(7/2),x)
Output:
(cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x)) ^(1/2)*(420*A*sin(c + d*x) + 490*B*sin(c + d*x) + 560*C*sin(c + d*x) + 126 *A*sin(2*c + 2*d*x) + 36*A*sin(3*c + 3*d*x) + 15*A*sin(4*c + 4*d*x) + 112* B*sin(2*c + 2*d*x) + 42*B*sin(3*c + 3*d*x) + 140*C*sin(2*c + 2*d*x)))/(420 *d*(cos(c + d*x) + 1))
\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2 ),x)
Output:
sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**4,x )*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**3,x)*b + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**2,x)*c)